Answer
$$\frac{2v^2+10v-48}{8v+64}=\frac{v-3}{4}$$
Explanation
| $$ \begin{aligned}\frac{2v^2+10v-48}{8v+64}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{v-3}{4}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{2v^2+10v-48}{8v+64} $ to $ \dfrac{v-3}{4} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2v+16}$. $$ \begin{aligned} \frac{2v^2+10v-48}{8v+64} & =\frac{ \left( v-3 \right) \cdot \color{blue}{ \left( 2v+16 \right) }}{ 4 \cdot \color{blue}{ \left( 2v+16 \right) }} = \\[1ex] &= \frac{v-3}{4} \end{aligned} $$ |
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Simplify Rational Expressions